Which of the limits is equivalent to the following definite integral? $ \int_0^5 (x+1)\,dx$ Choose 1 answer: Choose 1 answer: (Choice A) A $ \lim_{n\to\infty}\sum_{i=0}^n\left(\dfrac{5i-1}n+1\right)\cdot\dfrac{5}{n}$ (Choice B) B $ \lim_{n\to\infty}\sum_{i=0}^n\left(\dfrac{5i+1}n\right)\cdot\dfrac{5}{n}$ (Choice C) C $ \lim_{n\to\infty}\sum_{i=1}^n\left(\dfrac{5i}n+1\right)\cdot\dfrac{5}{n}$ (Choice D) D $ \lim_{n\to\infty}\sum_{i=1}^n\left(\dfrac{5i+1}n+1\right)\cdot\dfrac{5}{n}$
Solution: The value of the definite integral $ \int_0^5 (x+1)\,dx$ is the limit of its Riemann sums as the number of terms tends to infinity. These can be left Riemann sums, right Riemann sums, or some other kind of Riemann sums. For this definite integral, how wide are equal subintervals in a Riemann sum with $n$ terms? A Riemann sum with $n$ terms comes from dividing the interval of integration $[a,b]$ into $n$ subintervals. The given integral has $a=0$ and $b=5$. If we subdivide the interval $[0,5]$ into $n$ subintervals of equal width, we get a common width of $ \Delta x=\dfrac{b-a}n=\dfrac{5-0}n=\dfrac5n\,$. What are the left and right Riemann sums for the integral, expressed in terms of the integrand $f(x)=x+1$ ? The left Riemann sum with $n$ terms is $\begin{aligned} \sum_{i=0}^{n-1} f(a+i\Delta x)\cdot \Delta x &= \sum_{i=0}^{n-1} f\left(0+i\dfrac5n\right)\cdot \dfrac5n \\ \\ &= \sum_{i=0}^{n-1} f\left(\dfrac{5i}n\right)\cdot \dfrac5n \\ \\ &= \sum_{i=0}^{n-1} \left(\dfrac{5i}n+1\right)\cdot \dfrac5n \end{aligned}$ The right Riemann sum with $n$ terms is $ \sum_{i=1}^n f(a+i\Delta x)\cdot \Delta x = \sum_{i=1}^n\left(\dfrac{5i}n+1\right)\cdot \dfrac5n$ Do the answer choices include a left or right Riemann sum with $n$ terms? The answer choices include a right Riemann sum with $n$ terms. The correct answer is $ \int_0^5 (x+1)\,dx = \lim_{n\to\infty} \sum_{i=1}^n \left(\dfrac{5i}n+1\right)\cdot \dfrac5n$.